Answer:


Explanation:
Hello,
Clausius Clapeyron equation is suitable in this case, since it allows us to relate the P,T,V behavior along the described melting process and the associated energy change. Such equation is:

As both the enthalpy and volume do not change with neither the temperature nor the pressure for melting processes, its integration turns out:

Solving for the enthalpy of fusion we obtain:

Finally the entropy of fusion is given by:

Best regards.
The empirical formula : MnO₂.
<h3>Further explanation</h3>
Given
632mg of manganese(Mn) = 0.632 g
368mg of oxygen(O) = 0.368 g
M Mn = 55
M O = 16
Required
The empirical formula
Solution
You didn't include the pictures, but the steps for finding the empirical formula are generally the same
- Find mol(mass : atomic mass)
Mn : 0.632 : 55 = 0.0115
O : 0.368 : 16 =0.023
- Divide by the smallest mol(Mn=0.0115)
Mn : O =

The empirical formula : MnO₂
When you know the number of moles in either the reactant or product and also when the compounds involved are either basic or acidic
Answer:
<em>that</em><em> </em><em>obj</em><em>ect</em><em> </em><em>will</em><em> </em><em>float</em><em> </em><em>on</em><em> </em><em>the</em><em> </em><em>water</em>
Explanation:
Since the density of water is greater than that of the object , the object will float on the water.
NB. Density of the object=

Given that mass of the object is 80g and volume 100cm^3
substituting them in the formula
Density=80g/100cm^3
Density=0.8g/cm^3
<em>sin</em><em>ce</em><em> </em><em>0</em><em>.</em><em>8</em><em><</em><em>1</em><em> </em><em>whi</em><em>ch</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>dens</em><em>ity</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>obj</em><em>ect</em><em> </em><em>and</em><em> </em><em>water</em><em> </em><em>respect</em><em>ively</em><em>,</em><em> </em><em>the </em><em>obj</em><em>ect</em><em> </em><em>will</em><em> </em><em>float</em><em>.</em>
the concentration has been given as g/L. this is the mass of solute that can be dissolved in 1 L solution.
the concentration of the solution that needs to be prepared - 3.81 g/L
the mass of sodium nitrate is - 5.40 g
the volume needed for 3.18 g - 1 L
then the volume needed for 5.40 g - 1/3.18 x 5.40
volume of water needed - 1.42 L